# Derivative of Squareroot

What would be the derivate of square roots? For example if i have $2 \sqrt{x}$ or $\sqrt{x}$

Im unsure how to derive these and include them especially in something like implicit

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$\sqrt x=x^{1/2}$, so you just use the power rule: the derivative is $\frac12x^{-1/2}$.

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what about something like $2\sqrt{x}$ – soniccool Jun 29 '12 at 21:51
@soniccool: The derivative of $2f(x)$ is always twice the derivative of $f(x)$. So the derivative of $2\sqrt x$ is $2\cdot\frac12x^{-1/2}$. – MJD Jun 29 '12 at 21:52
@soniccool: You mean $\left(2\sqrt x\right)'=\frac1{\sqrt x}$. For the same reason that the derivative of $2x^3$ is $6x^2$: the derivative of $af(x)$ is $af'(x)$. In this case you have the derivative of $2x^{1/2}$, which is $$2\left(x^{1/2}\right)'=2\cdot\frac12x^{-1/2}=x^{-1/2}=\frac1{\sqrt x}\;.$$ – Brian M. Scott Jun 29 '12 at 22:19
Yes. Let $$y = 4\sqrt{x}$$ $$y' = 4(\frac{d}{dx} \sqrt{x})$$ $$y' = 4\cdot \frac{1}{2\sqrt{x}}$$ $$y' = \frac{2}{\sqrt{x}}$$ – Joe Jun 29 '12 at 22:49
@soniccool: You’re welcome. – Brian M. Scott Jun 29 '12 at 23:04

Let $f(x) = \sqrt{x}$, then $$f'(x) = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} \times \dfrac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \lim_{x \to 0} \dfrac{x+h-x}{h (\sqrt{x+h} + \sqrt{x})}\\ = \lim_{h \to 0} \dfrac{h}{h (\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \dfrac1{(\sqrt{x+h} + \sqrt{x})} = \dfrac1{2\sqrt{x}}$$ In general, you can use the fact that if $f(x) = x^{t}$, then $f'(x) = tx^{t-1}$.

Taking $t=1/2$, gives us that $f'(x) = \dfrac12 x^{-1/2}$, which is the same as we obtained above.

Also, recall that $\dfrac{d (c f(x))}{dx} = c \dfrac{df(x)}{dx}$. Hence, you can pull out the constant and then differentiate it.

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This is both exceedingly complete, and utterly useless for the OP. Good work! (+1) – Chris Taylor Jun 29 '12 at 22:01
This is the best answer here, because it doesn't assume that the power rule (which is easy to prove when the exponent is a positive integer) automatically applies when the exponent is NOT a positive integer. – user22805 Jun 29 '12 at 22:47

Let $f(x) = \sqrt{x} = x^{1/2}$.

$$f'(x) = \frac{1}{2} x ^{-1/2}$$

$$f'(x) = \frac{1}{2x^{1/2}} = \frac{1}{2\sqrt{x}}$$

If you post the specific implicit differentiation problem, it may help. The general guideline of writing the square root as a fractional power and then using the power and chain rule appropriately should be fine however. Also, remember that you can simply pull out a constant when dealing with derivatives - see below.

If $g(x) = 2\sqrt{x} = 2x^{1/2}$. Then,

$$g'(x) = 2\cdot\frac{1}{2}x^{-1/2}$$

$$g'(x) = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$$

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The Power Rule says that $\frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}$. Applying this to $\sqrt{x}=x^{\frac12}$ gives \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x} &=\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac12}\\ &=\frac12x^{-\frac12}\\ &=\frac{1}{2\sqrt{x}}\tag{1} \end{align} However, if you are uncomfortable applying the Power Rule to a fractional power, consider applying implicit differentiation to \begin{align} y&=\sqrt{x}\\ y^2&=x\\ 2y\frac{\mathrm{d}y}{\mathrm{d}x}&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{2y}\\ &=\frac{1}{2\sqrt{x}}\tag{2} \end{align}

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